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branch-name - Website Staff UI
SQL (Structured Query Language)
Source: Silbershatz, Fanny Santosa dan Aniati Murni
n Concepts of Query
n Basic Structure
n Set Operations
n Aggregate Functions
n Null Values
n Nested Subqueries
n Derived Relations
n Views
n Modification of the Database
n Joined Relations
Database System Concepts
4.1
©Silberschatz, Korth and Sudarshan
Tujuan
n Menjelaskan bahasa formal yang digunakan
pada basis data relasional
n Menjelaskan SQL sebagai bahasa yang
standar
n Memberikan beberapa contoh teknik
pembuatan query dengan menggunakan SQL
Database System Concepts
4.2
©Silberschatz, Korth and Sudarshan
Pengertian Query
n Query adalah perintah-perintah
untuk mengakses data pada sistem
basis data
Database System Concepts
4.3
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SQL (1)
n SQL adalah bahasa query baku untuk DBMS
n SQL diambil sebagai bakuan sejak tahun 1992
n Awalnya diterapkan pada DBMS besar seperti Oracle
dan Informix, sekarang juga pada DBMS berbasis PC
seperti dBASE dan FoxPro.
n SQL bersifat sebagai bahasa tingkat tinggi (high
level). Pemakai hanya menyebutkan hasil yang
diinginkan dan optimasi pelaksanaan query dilakukan
oleh DBMS.
n Satu perintah SQL dapat mewakili puluhan baris
perintah bahasa xBASE.
Database System Concepts
4.4
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SQL (2)
n SQL dapat disisipkan ke bahasa pemrograman yang lain seperti
C, Pascal, Cobol, dll.
n Bahasa SQL terbagi dalam dua bagian besar, yaitu: DDL (Data
Definition Language) dan DML (Data Manipulation Language)
n DDL mendefinisikan struktur basis data, seperti pembuatan
basis data, pembuatan tabel dsbnya. Contoh: CREATE
DATABASE dan CREATE TABLE.
n DML merupakan bagian untuk memanipulasi basis data seperti:
pengaksesan data, penghapusan, penambahan dan
pengubahan data. DML juga dapat digunakan untuk melakukan
komputasi data.
Contoh: INSERT,
DELETE, dan UPDATE.
Database System Concepts
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DDL (1)
n Perintah SQL untuk definisi data:
H CREATE untuk membentuk basis data, taable atau index
H ALTER untuk mengubah struktur table
H DROP untuk menghapus basis data, table atau index
n CREATE DATABASE
H Untuk membentuk basis data
H Sintaks: CREATE DATABASE nama_database
H Contoh: CREATE DATABASE COMPANY
n CREATE TABLE
H Untuk membentuk table dari basis data
H Untuk menyebutkan spesifikasi dan batasan atribut
Database System Concepts
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DDL (2)
n Contoh CREATE TABLE:
CREATE TABLE EMPLOYEE
(
PNAME
CHAR(15)
NOT NULL
LNAME
CHAR(15)
NOT NULL
SSN
CHAR(9)
NOT NULL
BDATE
DATE
ADDRESS
CHAR(30)
SEX
CHAR
SALARYDECIMAL(10.2)
DNO
Database System Concepts
CHAR(10)
4.7
);
©Silberschatz, Korth and Sudarshan
DDL (3)
n ALTER TABLE
HDigunakan untuk mengubah struktur table
HContoh kasus: misalkan ingin
menambahkan kolom JOB pada table
EMPLOYEE dengan tipe karakter selebar
12.
HPerintah:
ALTER TABLE EMPLOYEE ADD JOB CHAR(12);
Database System Concepts
4.8
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DDL (4)
n CREATE INDEX
H Membentuk berkas index dari table
H Index digunakan untuk mempercepat proses
pencarian
H Sintaks: CREATE [UNIQUE] INDEX nama_index
ON nama_table(kolom1, kolom2, …. )
H Contoh: CREATE INDEX EMPLOYEENDX ON
EMPLOYEE(SSN)
Database System Concepts
4.9
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DDL (5)
n Menghapus Basis Data
H DROP DATABASE
H Sintaks: DROP DATABASE nama_database
H Contoh: DROP DATABASE COMPANY
n Menghapus Table
H DROP TABLE
H Sintaks: DROP TABLE nama_table
H Contoh: DROP TABLE EMPLOYEE
n Menghapus Berkas Index
H DROP INDEX
H Sintaks: DROP INDEX nama_index
H Contoh: DROP INDEX EMPLOYEENDX
Database System Concepts
4.10
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DML (1)
n Bahasa untuk mengakses basis data
n Bahasa untuk mengolah basis data
n Bahasa untuk memanggil fungsi-fungsi agregasi
n Bahasa untuk melakukan query
n Jenis-jenis query:
H Sederhana
H Join
H Bertingkat
Database System Concepts
4.11
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Schema Used in Examples
Database System Concepts
4.12
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Basic Structure
n SQL is based on set and relational operations with certain
modifications and enhancements
n A typical SQL query has the form:
select A1, A2, ..., An
from r1, r2, ..., rm
where P
H Ais represent attributes
H ris represent relations
H P is a predicate.
n This query is equivalent to the relational algebra expression.
∏A1, A2, ..., An(σP (r1 x r2 x ... x rm))
n The result of an SQL query is a relation.
Database System Concepts
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Query Sederhana
n Bentuk umum SQL:
SELECT [ALL|DISTINCT]
nama_kolom_kolom_tabel
[INTO nama_tabel]
[FROM nama_nama_tabel]
[WHERE predikat]
[GROUP BY ekspresi]
[ORDER BY nama+kolom_tabel]
Database System Concepts
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The select Clause (1)
n The select clause corresponds to the projection operation of the relational
algebra. It is used to list the attributes desired in the result of a query.
n Find the names of all branches in the loan relation
select branch-name
from loan
n In the “pure” relational algebra syntax, the query would be:
∏branch-name(loan)
n An asterisk in the select clause denotes “all attributes”
select *
from loan
n NOTE: SQL does not permit the ‘-’ character in names, so you would use, for
example, branch_name instead of branch-name in a real implementation. We
use ‘-’ since it looks nicer!
n NOTE: SQL names are case insensitive, meaning you can use upper case or
lower case.
H You may wish to use upper case in places where we use bold font.
Database System Concepts
4.15
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The select Clause (2)
n SQL allows duplicates in relations as well as in query results.
n To force the elimination of duplicates (menghilangkan duplikasi penampilan
output yang sama) insert the keyword distinct after select.
Find the names of all branches in the loan relations, and remove duplicates
select distinct branch-name
from loan
n The keyword all specifies that duplicates not be removed.
select all branch-name
from loan
n Menampilkan isi tabel customer
select * from customer
n Menampilkan semua fname dari tabel customer
select customer_name from customer
Database System Concepts
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The select Clause (3)
n Menampilkan account number dan balance dari kantor cabang
(branch_name) “Pondok Kelapa”
select account_number, balance
from account
where branch_name = “Pondok Kelapa”;
n Perintah diatas dapat juga dituliskan dengan menggunakan qualified
column names sebagai berikut:
select account.account_number,
account.balance
from account
where branch_name = “Pondok Kelapa”;
Database System Concepts
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The select Clause (4)
n The select clause can contain arithmetic expressions involving the
operation, +, –, ∗, and /, and operating on constants or attributes of
tuples.
n The query:
select loan-number, branch-name, amount ∗ 100
from loan
would return a relation which is the same as the loan relations, except
that the attribute amount is multiplied by 100.
Database System Concepts
4.18
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The where Clause (1)
n The where clause corresponds to the selection predicate of the
relational algebra. If consists of a predicate involving attributes of the
relations that appear in the from clause.
n The find all loan number for loans made a the Perryridge branch with
loan amounts greater than $1200.
select loan-number
from loan
where branch-name = ‘Perryridge’ and amount > 1200
n Comparison results can be combined using the logical connectives and,
or, and not.
n Comparisons can be applied to results of arithmetic expressions.
Database System Concepts
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The where Clause (2)
n SQL Includes a between comparison operator in order to simplify where
clauses that specify that a value be less than or equal to some value and
greater than or equal to some other value.
n Find the loan number of those loans with loan amounts between $90,000
and $100,000 (that is, ≥$90,000 and ≤$100,000)
select loan-number
from loan
where amount between 90000 and 100000
Database System Concepts
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The from Clause – Contoh Query Join
n The from clause corresponds to the Cartesian product operation of the relational
algebra. It lists the relations to be scanned in the evaluation of the expression.
n Find the Cartesian product borrower x loan
select ∗
from borrower, loan
n Find the name, loan number and loan amount of all customers having a loan at
the Perryridge branch.
select customer-name, borrower.loan-number, amount
from borrower, loan
where borrower.loan-number = loan.loan-number and
branch-name = ‘Perryridge’
n Contoh diatas merupakan Query Join dengan
branch_name =
“Perryridge” sebagai kondisi seleksi dan borrower.loan-number = loan.loannumber sebagai kondisi join
Database System Concepts
4.21
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The Rename Operation
n The SQL allows renaming relations and attributes using the as clause:
old-name as new-name
n Find the name, loan number and loan amount of all customers; rename the
column name loan-number as loan-id.
select customer-name, borrower.loan-number as loan-id, amount
from borrower, loan
where borrower.loan-number = loan.loan-number
Database System Concepts
4.22
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String Operations
n SQL includes a string-matching operator for comparisons on character
strings.
n Find the names of all customers whose street includes the substring
“Main”.
select customer-name
from customer
where customer-street like ‘%Main%’
n SQL supports a variety of string operations such as
H concatenation (using “||”)
H converting from upper to lower case (and vice versa)
H finding string length, extracting substrings, etc.
Database System Concepts
4.23
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Ordering the Display of Tuples
n List in alphabetic order the names of all customers having a loan in
Perryridge branch
select distinct customer-name
from borrower, loan
where borrower.loan-number = loan.loan-number and
branch-name = ‘Perryridge’
order by customer-name
n We may specify desc for descending order or asc for ascending order,
for each attribute; ascending order is the default.
H E.g. order by customer-name desc
Database System Concepts
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Set Operations (1)
n The set operations union, intersect, and except operate on relations
and correspond to the relational algebra operations ∪, ∩, −.
n Each of the above operations automatically eliminates duplicates; to
retain all duplicates use the corresponding multiset versions union all,
intersect all and except all.
Suppose a tuple occurs m times in r and n times in s, then, it occurs:
H m + n times in r union all s
H min(m,n) times in r intersect all s
H max(0, m – n) times in r except all s
Database System Concepts
4.25
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Set Operations (2)
n Find all customers who have a loan, an account, or both:
(select customer-name from depositor)
union
(select customer-name from borrower)
n Find all customers who have both a loan and an account.
(select customer-name from depositor)
intersect
(select customer-name from borrower)
n Find all customers who have an account but no loan.
(select customer-name from depositor)
except
(select customer-name from borrower)
Database System Concepts
4.26
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Aggregate Functions (1)
n These functions operate on the multiset of values of a column of a
relation, and return a value
avg: average value
min: minimum value
max: maximum value
sum: sum of values
count: number of values
Database System Concepts
4.27
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Aggregate Functions (2)
n Find the average account balance at the Perryridge branch.
select avg (balance)
from account
where branch-name = ‘Perryridge’
n Find the number of tuples in the customer relation.
select count (*)
from customer
n Find the number of depositors in the bank.
select count (distinct customer-name)
from depositor
Database System Concepts
4.28
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Aggregate Functions – Group By
n Find the number of depositors for each branch.
select branch-name, count (distinct customer-name)
from depositor, account
where depositor.account-number = account.account-number
group by branch-name
Note: Attributes in select clause outside of aggregate functions must
appear in group by list
Database System Concepts
4.29
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Aggregate Functions – Having Clause
n Find the names of all branches where the average account balance is
more than $1,200.
select branch-name, avg (balance)
from account
group by branch-name
having avg (balance) > 1200
Note: predicates in the having clause are applied after the formation of
groups whereas predicates in the where clause are applied before
forming groups
Database System Concepts
4.30
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Null Values (1)
n It is possible for tuples to have a null value, denoted by null, for some
of their attributes
n null signifies an unknown value or that a value does not exist.
n The predicate is null can be used to check for null values.
H E.g. Find all loan number which appear in the loan relation with null values
for amount.
select loan-number
from loan
where amount is null
n The result of any arithmetic expression involving null is null
H E.g. 5 + null returns null
n However, aggregate functions simply ignore nulls
H more on this shortly
Database System Concepts
4.31
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Null Values and Three Valued Logic (2)
n Any comparison with null returns unknown
H E.g. 5 < null or null <> null or null = null
n Three-valued logic using the truth value unknown:
H OR: (unknown or true) = true, (unknown or false) = unknown
(unknown or unknown) = unknown
H AND: (true and unknown) = unknown, (false and unknown) = false,
(unknown and unknown) = unknown
H NOT: (not unknown) = unknown
H “P is unknown” evaluates to true if predicate P evaluates to unknown
n Result of where clause predicate is treated as false if it evaluates to
unknown
Database System Concepts
4.32
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Null Values and Aggregates (3)
n Total all loan amounts
select sum (amount)
from loan
H Above statement ignores null amounts
H result is null if there is no non-null amount, that is the
n All aggregate operations except count(*) ignore tuples with null values
on the aggregated attributes.
Database System Concepts
4.33
©Silberschatz, Korth and Sudarshan
Nested Subqueries – Query Bertingkat
n SQL provides a mechanism for the nesting of subqueries.
n A subquery is a select-from-where expression that is nested within another
query.
n A common use of subqueries is to perform tests for set membership, set
comparisons, and set cardinality.
n Find all customers who have both an account and a loan at the bank.
select distinct customer-name
from borrower
where customer-name in (select customer-name
from depositor)
n Find all customers who have a loan at the bank but do not have an account at
the bank
select distinct customer-name
from borrower
where customer-name not in (select customer-name
from depositor)
Database System Concepts
4.34
©Silberschatz, Korth and Sudarshan
Example Nested Query
n Find all customers who have both an account and a loan at the
Perryridge branch
select distinct customer-name
from borrower, loan
where borrower.loan-number = loan.loan-number and
branch-name = “Perryridge” and
(branch-name, customer-name) in
(select branch-name, customer-name
from depositor, account
where depositor.account-number =
account.account-number)
n Note: Above query can be written in a much simpler manner. The
formulation above is simply to illustrate SQL features.
(Schema used in this example)
Database System Concepts
4.35
©Silberschatz, Korth and Sudarshan
Set Comparison and Tuple Variables
n Find all branches that have greater assets than some branch located in
Brooklyn.
select distinct T.branch-name
from branch as T, branch as S
where T.assets > S.assets and
S.branch-city = ‘Brooklyn’
n Same query using > some clause
select branch-name
from branch
where assets > some
(select assets
from branch
where branch-city = ‘Brooklyn’)
Database System Concepts
4.36
©Silberschatz, Korth and Sudarshan
Definition of Some Clause Existential Quantifier
n F <comp> some r ⇔ ∃ t ∈ r s.t. (F <comp> t)
Where <comp> can be: <, ≤, >, =, ≠
0
5
6
) = true
(5< some
0
5
) = false
(5 = some
0
5
) = true
(5 ≠ some
0
5
) = true (since 0 ≠ 5)
(5< some
(read: 5 < some tuple in the relation)
(= some) ≡ in
However, (≠ some) ≡ not in
Database System Concepts
4.37
©Silberschatz, Korth and Sudarshan
Definition of all Clause
n F <comp> all r ⇔ ∀ t ∈ r (F <comp> t)
(5< all
0
5
6
) = false
(5< all
6
10
) = true
(5 = all
4
5
) = false
(5 ≠ all
4
6
) = true (since 5 ≠ 4 and 5 ≠ 6)
(≠ all) ≡ not in
However, (= all) ≡ in
Database System Concepts
4.38
©Silberschatz, Korth and Sudarshan
Example Query
n Find the names of all branches that have greater assets than all
branches located in Brooklyn.
select branch-name
from branch
where assets > all
(select assets
from branch
where branch-city = ‘Brooklyn’)
Database System Concepts
4.39
©Silberschatz, Korth and Sudarshan
Test for Empty Relations
n The exists construct returns the value true if the argument subquery is
nonempty.
n exists r ⇔ r ≠ Ø
n not exists r ⇔ r = Ø
Database System Concepts
4.40
©Silberschatz, Korth and Sudarshan
Example Query
n Find all customers who have an account at all branches located in Brooklyn.
select distinct S.customer-name
from depositor as S
where not exists (
(select branch-name
from branch
where branch-city = ‘Brooklyn’)
except
(select R.branch-name
from depositor as T, account as R
where T.account-number = R.account-number and
S.customer-name = T.customer-name))
n (Schema used in this example)
n Note that X – Y = Ø ⇔ X ⊆ Y
n Note: Cannot write this query using = all and its variants
Database System Concepts
4.41
©Silberschatz, Korth and Sudarshan
Test for Absence of Duplicate Tuples
n The unique construct tests whether a subquery has any duplicate
tuples in its result.
n Find all customers who have at most one account at the Perryridge
branch.
select T.customer-name
from depositor as T
where unique (
select R.customer-name
from account, depositor as R
where T.customer-name = R.customer-name and
R.account-number = account.account-number and
account.branch-name = ‘Perryridge’)
n (Schema used in this example)
Database System Concepts
4.42
©Silberschatz, Korth and Sudarshan
Example Query
n Find all customers who have at least two accounts at the Perryridge
branch.
select distinct T.customer-name
from depositor T
where not unique (
select R.customer-name
from account, depositor as R
where T.customer-name = R.customer-name and
R.account-number = account.account-number and
account.branch-name = ‘Perryridge’)
n (Schema used in this example)
Database System Concepts
4.43
©Silberschatz, Korth and Sudarshan
Views
n Provide a mechanism to hide certain data from the view of certain users.
To create a view we use the command:
create view v as <query expression>
where:
H <query expression> is any legal expression
H The view name is represented by v
Database System Concepts
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©Silberschatz, Korth and Sudarshan
Example Queries
n A view consisting of branches and their customers
create view all-customer as
(select branch-name, customer-name
from depositor, account
where depositor.account-number = account.account-number)
union
(select branch-name, customer-name
from borrower, loan
where borrower.loan-number = loan.loan-number)
n Find all customers of the Perryridge branch
select customer-name
from all-customer
where branch-name = ‘Perryridge’
Database System Concepts
4.45
©Silberschatz, Korth and Sudarshan
Derived Relations
n Find the average account balance of those branches where the average
account balance is greater than $1200.
select branch-name, avg-balance
from (select branch-name, avg (balance)
from account
group by branch-name)
as result (branch-name, avg-balance)
where avg-balance > 1200
Note that we do not need to use the having clause, since we compute
the temporary (view) relation result in the from clause, and the
attributes of result can be used directly in the where clause.
Database System Concepts
4.46
©Silberschatz, Korth and Sudarshan
With Clause
n With clause allows views to be defined locally to a query, rather than
globally. Analogous to procedures in a programming language.
n Find all accounts with the maximum balance
with max-balance(value) as
select max (balance)
from account
select account-number
from account, max-balance
where account.balance = max-balance.value
Database System Concepts
4.47
©Silberschatz, Korth and Sudarshan
Complex Query using With Clause (PR)
n Find all branches where the total account deposit is greater than the
average of the total account deposits at all branches
with branch-total (branch-name, value) as
select branch-name, sum (balance)
from account
group by branch-name
with branch-total-avg(value) as
select avg (value)
from branch-total
select branch-name
from branch-total, branch-total-avg
where branch-total.value >= branch-total-avg.value
Database System Concepts
4.48
©Silberschatz, Korth and Sudarshan
Modification of the Database – Deletion
n Delete all account records at the Perryridge branch
delete from account
where branch-name = ‘Perryridge’
n Delete all accounts at every branch located in Needham city.
delete from account
where branch-name in (select branch-name
from branch
where branch-city = ‘Needham’)
delete from depositor
where account-number in
(select account-number
from branch, account
where branch-city = ‘Needham’
and branch.branch-name = account.branch-name)
n (Schema used in this example)
Database System Concepts
4.49
©Silberschatz, Korth and Sudarshan
Example Query
n Delete the record of all accounts with balances below the average at the
bank.
delete from account
where balance < (select avg (balance)
from account)
H Problem: as we delete tuples from deposit, the average balance changes
H Solution used in SQL:
1. First, compute avg balance and find all tuples to delete
2. Next, delete all tuples found above (without recomputing avg or retesting
the tuples)
Database System Concepts
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©Silberschatz, Korth and Sudarshan
ModificationoftheDatabase – Insertion
n Add a new tuple to account
insert into account
values (‘A-9732’, ‘Perryridge’,1200)
or equivalently
insert into account (branch-name, balance, account-number)
values (‘Perryridge’, 1200, ‘A-9732’)
n Add a new tuple to account with balance set to null
insert into account
values (‘A-777’,‘Perryridge’, null)
Database System Concepts
4.51
©Silberschatz, Korth and Sudarshan
Modification of the Database – Insertion
n Provide as a gift for all loan customers of the Perryridge branch, a $200
savings account. Let the loan number serve as the account number for
the new savings account
insert into account
select loan-number, branch-name, 200
from loan
where branch-name = ‘Perryridge’
insert into depositor
select customer-name, loan-number
from loan, borrower
where branch-name = ‘Perryridge’
and loan.account-number = borrower.account-number
n The select from where statement is fully evaluated before any of its
results are inserted into the relation (otherwise queries like
insert into table1 select * from table1
would cause problems
Database System Concepts
4.52
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Modification of the Database – Updates
n Increase all accounts with balances over $10,000 by 6%, all other
accounts receive 5%.
H Write two update statements:
update account
set balance = balance ∗ 1.06
where balance > 10000
update account
set balance = balance ∗ 1.05
where balance ≤ 10000
H The order is important
H Can be done better using the case statement (next slide)
Database System Concepts
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©Silberschatz, Korth and Sudarshan
Case Statement for Conditional Updates
n Same query as before: Increase all accounts with balances over
$10,000 by 6%, all other accounts receive 5%.
update account
set balance = case
when balance <= 10000 then balance *1.05
else balance * 1.06
end
Database System Concepts
4.54
©Silberschatz, Korth and Sudarshan
Update of a View
n Create a view of all loan data in loan relation, hiding the amount attribute
create view branch-loan as
select branch-name, loan-number
from loan
n Add a new tuple to branch-loan
insert into branch-loan
values (‘Perryridge’, ‘L-307’)
This insertion must be represented by the insertion of the tuple
(‘L-307’, ‘Perryridge’, null)
into the loan relation
n Updates on more complex views are difficult or impossible to translate, and
hence are disallowed.
n Most SQL implementations allow updates only on simple views (without
aggregates) defined on a single relation
Database System Concepts
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Transactions (1)
n A transaction is a sequence of queries and update statements executed as a
single unit
H Transactions are started implicitly and terminated by one of
4 commit work: makes all updates of the transaction permanent in the database
4 rollback work: undoes all updates performed by the transaction.
n Motivating example
H Transfer of money from one account to another involves two steps:
4 deduct from one account and credit to another
H If one steps succeeds and the other fails, database is in an inconsistent state
H Therefore, either both steps should succeed or neither should
n If any step of a transaction fails, all work done by the transaction can be undone
by rollback work.
n Rollback of incomplete transactions is done automatically, in case of system
failures
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Transactions (2)
n In most database systems, each SQL statement that executes
successfully is automatically committed.
H Each transaction would then consist of only a single statement
H Automatic commit can usually be turned off, allowing multi-statement
transactions, but how to do so depends on the database system
H Another option in SQL:1999: enclose statements within
begin atomic
…
end
Database System Concepts
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Joined Relations (PR)
n Join operations take two relations and return as a result another
relation.
n These additional operations are typically used as subquery expressions
in the from clause
n Join condition – defines which tuples in the two relations match, and
what attributes are present in the result of the join.
n Join type – defines how tuples in each relation that do not match any
tuple in the other relation (based on the join condition) are treated.
Join Types
Join Conditions
inner join
left outer join
right outer join
full outer join
natural
on <predicate>
using (A1, A2, ..., An)
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Joined Relations – Datasets for Examples
n Relation loan
loan-number
branch-name
amount
L-170
Downtown
3000
L-230
Redwood
4000
L-260
Perryridge
1700
n Relation borrower
customer-name
loan-number
Jones
L-170
Smith
L-230
Hayes
L-155
n Note: borrower information missing for L-260 and loan
information missing for L-155
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Joined Relations – Examples
n loan inner join borrower on
loan.loan-number = borrower.loan-number
loan-number
branch-name
amount
customer-name
loan-number
L-170
Downtown
3000
Jones
L-170
L-230
Redwood
4000
Smith
L-230
loan left inner join borrower on
loan.loan-number = borrower.loan-number
loan-number
branch-name
amount
customer-name
loan-number
L-170
Downtown
3000
Jones
L-170
L-230
Redwood
4000
Smith
L-230
L-260
Perryridge
1700
null
Database System Concepts
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Joined Relations – Examples
n loan natural inner join borrower
loan-number
branch-name
amount
customer-name
L-170
Downtown
3000
Jones
L-230
Redwood
4000
Smith
loan natural right outer join borrower
loan-number
Database System Concepts
branch-name
amount
customer-name
L-170
Downtown
3000
Jones
L-230
Redwood
4000
Smith
L-155
null
null
Hayes
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Joined Relations – Examples
n loan full outer join borrower using (loan-number)
loan-number
branch-name
amount
customer-name
L-170
Downtown
3000
Jones
L-230
Redwood
4000
Smith
L-260
Perryridge
1700
null
L-155
null
null
Hayes
Find all customers who have either an account or a loan (but not
both) at the bank.
select customer-name
from (depositor natural full outer join borrower)
where account-number is null or loan-number is null
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